During my Calculus course, my professor used a property to determine if a piecewise function is differentiable (and calculate the derivative of it) but it has never been proved. Here I attempt to reproduce a coherent statement for that property
Theorem Let $a, b, c \in \mathbb{R} : a < b < c$, let $f: (a, b] \to \mathbb{R}, g: [b, c) \to \mathbb{R}$ be differentiable functions respectively in $(a, b)$ and in $(b, c)$. Let $h: (a, c) \to \mathbb{R}$ such that $$ h(x) = \begin{cases} f(x) & \text{if } a < x \leq b\\ g(x) & \text{if } b < x < c \end{cases} \, , $$ if $f(b) = g(b)$ and $\lim_{x \to b^-} f'(x) = \lim_{x \to b^+} g'(x) = l$, then $h$ is differentiable in $(a, c)$ and $h'(b) = l$.
I think that it can be reformulated like this
Theorem Let $a, b, c \in \mathbb{R} : a < b < c$, let $f: (a, c) \to \mathbb{R}$ be a continuous function in $b$, differentiable in $(a, c)\setminus\{b\}$. If $\lim_{x \to b^-} f'(x) = \lim_{x \to b^+} f'(x) = l$ then $f$ is differentiable in $b$ and $f'(b) = l$.
Proof I want to prove separately that $$ \lim_{x \to b^-} \frac{f(b) - f(x)}{b - x} = l \, , \quad \lim_{x \to b^+} \frac{f(b) - f(x)}{b - x} = l \, . $$ Let $\varepsilon > 0$, due to the fact that $\lim_{x \to b^-} f'(x) = l$, we have $\exists \, \delta > 0$ such that $|l - f'(x)| < \varepsilon \ \forall x \in \mathbb{R} : b - \delta < x < b$. Let $b - \delta < x < b$ then $$\left| \frac{f(b) - f(x)}{b - x} - l \right| \leq \left| \frac{f(b) - f(x)}{b - x} - f'(x) \right| + |f'(x) - l| \leq \left| \frac{f(b) - f(x)}{b - x} - f'(x) \right| + \varepsilon$$
I don't know how to "stimate" $\left| \frac{f(b) - f(x)}{b - x} - f'(x) \right|$ using $\varepsilon$, so that I can show the limit is correct.
I would like to know if I'm missing some hypotheses and how this theorem (or the "corrected one") can be proved.
Let $\varepsilon >0$. Due to the fact that $\lim_{x\to b-}f′(x)=l$, we have $\delta>0$ such that $|l−f′(x)|<\varepsilon$ for all $ b−\delta<x<b$. Take $b−\delta<x<b$. Then $\frac{f(b) - f(x)}{b - x} = f'(c)$ for some $c\in (x,b)\subset (b-\delta,b)$. Therefore $$ \left|\frac{f(b) - f(x)}{b - x} - l\right|= |f'(c)-l|<\varepsilon. $$